RD Chapter 1 Real Numbers Ex MCQS Solutions
Question - 21 : - If n is a natural number, then 92n – 42n isalways divisible by
(a) 5
(b) 13
(c) both 5 and 13
(d) None of these
Answer - 21 : - (c)
Question - 22 : - If n is any natural number, then 6n – 5n always ends with
(a) 1
(b) 3
(c) 5
(d) 7
Answer - 22 : -
(a) nis any natural number and 6n – 5n
We know that 6n ends with 6 and 5n ends with 5
6n – 5n will end with 6 – 5 = 1
Question - 23 : - The LCM and HCF of two rational numbers are equal, then the numbers must be
(a) prime
(b) co-prime
(c) composite
(d) equal
Answer - 23 : - (d) LCM and HCF of two rational numbers are equal Then those must be equal
Question - 24 : - If the sum of LCM and HCF of two numbers is 1260 and their LCM is 900 more than their HCF, then the product of two numbers is
(a) 203400
(b) 194400
(c) 198400
(d) 205400
Answer - 24 : -
(b) Sum of LCM and HCF of two numbers = 1260
LCM = 900 more than HCF
LCM = 900 +HCF
But LCM = HCF = 1260
900 + HCF + HCF = 1260
=> 2HCF = 1260 – 900 = 360
=> HCF = 180
and LCM = 1260 – 180 = 1080
Product = LCM x HCF = 1080 x 180 = 194400
Product of numbers = 194400
Question - 25 : - The remainder when the square of any prime number greater than 3 is divided by 6, is
(a) 1
(b) 3
(c) 2
(d) 4
Answer - 25 : - (a)
Question - 26 : - For some integer m, every even integer is of the form
(a) m
(b) m + 1
(c) 2m
(d) 2m + 1
Answer - 26 : -
(c) We know that, even integers are 2, 4, 6, …
So, it can be written in the form of 2m Where, m = Integer = Z
[Since, integer is represented by Z]
or m = …, -1, 0, 1, 2, 3, …
2m = …, -2, 0, 2, 4, 6, …
Alternate Method
Let ‘a’ be a positive integer.
On dividing ‘a’ by 2, let m be the quotient and r be the remainder.
Then, by Euclid’s division algorithm, we have
a – 2m + r, where a ≤ r < 2 i.e., r = 0 and r = 1
=> a = 2 m or a = 2m + 1
When, a = 2m for some integer m, then clearly a is even.
Question - 27 : - For some integer q, every odd integer is of the form
(a) q
(b) q + 1
(c) 2q
(d) 2q + 1
Answer - 27 : -
(d) We know that, odd integers are 1, 3, 5,…
So, it can be written in the form of 2q + 1 Where, q = integer = Z
or q = …, -1, 0, 1, 2, 3, …
2q + 1 = …, -3, -1, 1, 3, 5, …
Alternate Method
Let ‘a’ be given positive integer.
On dividing ‘a’ by 2, let q be the quotient and r be the remainder.
Then, by Euclid’s division algorithm, we have
a = 2q + r, where 0 ≤ r < 2
=> a = 2q + r, where r = 0 or r = 1
=> a = 2q or 2q + 1
When a = 2q + 1 for some integer q, then clearly a is odd.
Question - 28 : - n2 – 1 is divisible by 8, if n is
(a) an integer
(b) a natural number
(c) an odd integer
(d) an even integer
Answer - 28 : - (c)
At k = -1, a = 4(-1)(-1 + 1) = 0 which is divisible by 8.
At k = 0, a = 4(0)(0 + 1) = 4 which is divisible by 8.
At k = 1, a = 4(1)(1 + 1) = 8 which is divisible by 8.
Hence, we can conclude from above two cases, if n is odd, then n2 –1 is divisible by 8.
(a) one decimal place
(b) two decimal places .
(c) three decimal places
(d) more than 3 decimal places
Answer - 29 : - (b)
Question - 30 : - If two positive integers a and b are written as a = x3y2 andb = xy3 ; x, y are prime numbers, then HCF (a, b) is
(a) xy
(b) xy2
(c) x3y3
(d) x2y2
Answer - 30 : - (b)
